Nondisjunction Math (Practice Q's spoiler?)

[Wererew]

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This might be confusing but please help me clarify

I understand that nondisjunction @anaphase I = either n+1 or n-1, @anaphase II = either n-1, n+1, or n

So possibilities for human gametes=

@ anaphase I = 22 or 24
@ anaphase II = 22, 23, or 24

So far I believe this is how it would go for a human gamete..


A question (from BC) asks for possibility for nondisjunction occurring @ anaphase I for an oocyte/egg, it will be 2n-1 or 2n+1

possibilities= 45, 47

However I am now confused about some concepts.

The first part with n-1,n+1 is AFTER the nondisjunction AND the whole meiosis completed, describing the resulting haploid...

But this BC question is asking what the resulting fertilized diploid number will be?

Would this not matter because two gametes will always be fusing with a whole number of haploid chromosomes, regardless of possible nondisjunction from the gamete?

 

[FeralisExtremum]

Would this not matter because two gametes will always be fusing with a whole number of haploid chromosomes, regardless of possible nondisjunction from the gamete?

I'm a bit unclear on your question here. It would matter if one of your gametes had a lower (or higher) haploid chromosome number before fusing with another gamete - even if the second gamete has a normal chromosome number (23), the other gamete is adding either too many or too few chromosomes as a result of the nondisjunction event (22 or 24). The result will therefore be a diploid zygote with either too few or too many chromosomes (45 or 47).

 

[Wererew]

I'm a bit unclear on your question here. It would matter if one of your gametes had a lower (or higher) haploid chromosome number before fusing with another gamete - even if the second gamete has a normal chromosome number (23), the other gamete is adding either too many or too few chromosomes as a result of the nondisjunction event (22 or 24). The result will therefore be a diploid zygote with either too few or too many chromosomes (45 or 47).

I guess my question is if two nondisjunction gametes both failure at anaphase 1 fused, wouldn't the possibility also be 44? (from n-1 and n-1 gametes)

So possibilities would be : 44, 45, 47,48?


I am not understanding how for the gametes, it is n-1 or n+1 but for the zygote it is 2n-1 or 2n+1, my math does not seem to match up. am I overthinking

 

[FeralisExtremum]

I guess my question is if two nondisjunction gametes both failure at anaphase 1 fused, wouldn't the possibility also be 44? (from n-1 and n-1 gametes)

So possibilities would be : 44, 45, 47,48?


I am not understanding how for the gametes, it is n-1 or n+1 but for the zygote it is 2n-1 or 2n+1, my math does not seem to match up. am I overthinking

I think you're including the possibility of both gametes (the sperm AND the egg) having nondisjunction events. The question you are referring to specifies that the nondisjunction event occurs in the oocyte (egg) - not the sperm - before asking what the chromosome number in the zygote will be. You assume the other gamete (the sperm) is normal by default.

 

[Wererew]

I think you're including the possibility of both gametes (the sperm AND the egg) having nondisjunction events. The question you are referring to specifies that the nondisjunction event occurs in the oocyte (egg) - not the sperm - before asking what the chromosome number in the zygote will be. You assume the other gamete (the sperm) is normal by default.

Ohhh okay I get it! Thanks a bunch!!